最常见的数组元素

Question

我需要在数组中找到最常见的(模态)元素.

我能想到的最简单的方法是为每个唯一元素设置变量,并为每个元素分配一个计数变量,每次在一个贯穿数组的for循环中记录它时,它都会增加.

不幸的是,数组的大小是未知的并且将非常大,因此这种方法是无用的.

我在Objective-C中遇到了一个类似的问题,它使用NSCountedSet方法对数组元素进行排名.不幸的是,我对编程很新,只能将第一行翻译成Swift.

建议的方法如下:

    var yourArray: NSArray! // My swift translation

    NSCountedSet *set = [[NSCountedSet alloc] initWithArray:yourArray];

    NSMutableDictionary *dict=[NSMutableDictionary new];

    for (id obj in set) {
        [dict setObject:[NSNumber numberWithInteger:[set countForObject:obj]]
            forKey:obj]; //key is date
    }

    NSLog(@"Dict : %@", dict);

    NSMutableArray *top3=[[NSMutableArray alloc]initWithCapacity:3];

    //which dict obj is = max
    if (dict.count>=3) {

        while (top3.count<3) {
            NSInteger max = [[[dict allValues] valueForKeyPath:@"@max.intValue"] intValue];

            for (id obj in set) {
                if (max == [dict[obj] integerValue]) {
                    NSLog(@"--> %@",obj);
                    [top3 addObject:obj];
                    [dict removeObjectForKey:obj];
                }
            }
        }
    }

    NSLog(@"top 3 = %@", top3);

在我的程序中,我需要在数组中找到前五个地名.

Answer 1

编辑:现在使用下面的Swift 2.0

不是最有效的解决方案,而是简单的解决方案:

let a = [1,1,2,3,1,7,4,6,7,2]

var frequency: [Int:Int] = [:]

for x in a {
    // set frequency to the current count of this element + 1
    frequency[x] = (frequency[x] ?? 0) + 1
}

let descending = sorted(frequency) { $0.1 > $1.1 }

descending现在由一对数组组成:值和频率,首先排序最频繁.所以"前5名"将是前5个条目(假设有5个或更多不同的值).源阵列有多大并不重要.

这是一个适用于任何序列的通用函数版本:

func frequencies
  <S: SequenceType where S.Generator.Element: Hashable>
  (source: S) -> [(S.Generator.Element,Int)] {

    var frequency: [S.Generator.Element:Int] = [:]

    for x in source {
        frequency[x] = (frequency[x] ?? 0) + 1
    }

    return sorted(frequency) { $0.1 > $1.1 }
}

frequencies(a)

对于Swift 2.0,您可以将该功能调整为协议扩展:

extension SequenceType where Generator.Element: Hashable {
    func frequencies() -> [(Generator.Element,Int)] {

        var frequency: [Generator.Element:Int] = [:]

        for x in self {
            frequency[x] = (frequency[x] ?? 0) + 1
        }

        return frequency.sort { $0.1 > $1.1 }
    }
}

a.frequencies()

对于Swift 3.0:

extension Sequence where Self.Iterator.Element: Hashable {
    func frequencies() -> [(Self.Iterator.Element,Int)] {

        var frequency: [Self.Iterator.Element:Int] = [:]

        for x in self {
            frequency[x] = (frequency[x] ?? 0) + 1
        }

        return frequency.sorted { $0.1 > $1.1 }
    }
}